# Pressure in a bomb calorimeter at the moment of combustion

Seibtsantos
Homework Statement:
The calorific value of food can be determined in a bomb calorimeter, which consists of a hermetically sealed stainless steel container, in which the sample is burned in the presence of pure oxygen and the resulting heat measured by the temperature variation of a water bath surrounding the container. Due to the thickness and mass of the container, the water bath undergoes a few degrees variation, however at the time of combustion the internal temperature can reach 1200°C.
In determining the heat of combustion of a glucose sample (C6H12O6), 1.8 g of this substance was placed in a calorimetric pump and the container (1 L capacity) was pressurized to 30 atm with pure oxygen (1.23 moles ) and the ignition perpetrated.
Consider: The adiabatic system, a complete combustion at 1200 °C and that the gases generated are ideal gases. Data: R = 0.082 atm.L.K-1.mol-1; Molar Mass (C6H12O6) = 180 g/mol.
The internal pressure of the container at the exact moment of the sample combustion is:
a) 124 atm.
b) 145 atm.
►c) 152 atm.
d) 163 atm.
e) 173 atm.
Relevant Equations:
Ideal gas law
First, I calculated the number of moles of glucose.

n = m / M
n = 1.8 / 180
n = 0.01 moles of glucose

So I checked the combustion reaction.

1 C6H12O6 + 6 O2 -> 6 CO2 + 6 H20
1 + 6 -> 6 + 6
0.01 + 0.06 -> 0.06 + 0.06

I considered the number of moles at the end of the reaction.
I subtracted the number of moles from the initial oxygen by the oxygen consumed.

1.23 - 0.06 = 1.17 moles

I calculated the total number of moles.

0.06 + 0.06 + 1.17 = 1.29 moles

So I used it in the equation.

pV = nRT
p = (nRT) / V
p = (1.29 * 0.082 * 1473) / 1 (considering 1200 ° C = 1473K)
p = 155.8 atm

I didn't get to the result and I don't know what is missing

Thank you!